Equilibrium Practice Problems 1. Write the equilibrium expression for each of the following reactions: N2 (g) + 3 H2 (g) 2 NH3 (g) I2 (s) + Cl2 (g) 2 ICl NO2 (g) NO (g) (g) + ½ O2 (g) 2. The dissociation of acetic acid, CH3COOH, has an equilibrium constant at 25C of 1.8 x 10-5. The reaction is CH3COOH (aq) CH3COO- (aq) + H+ (aq) If the equilibrium concentration of CH3COOH is 0.46 moles in 0.500 L of water and that of CH3COO- is 8.1 x 10-3 moles in the same 0.500 L, calculate [H+] for the reaction. 3. The reaction 2 NO (g) N2 (g) + O2 (g) has a value of K= 2400 at 2000 K. If 0.61 g of NO are put in a previously empty 3.00 L vessel, calculate the equilibrium concentrations of NO, N2, and O2. 4. The following reaction has an equilibrium constant of 620 at a certain temperature. Calculate the equilibrium concentrations of all species if 4.5 mol of each component were added to a 3.0 L flask. H2 (g) + F2 (g) 2 HF (g)\ 5. Ammonia undergoes hydrolysis according to the following reaction: NH3 (aq) + H2O NH4+ (aq) + OH- (aq) K = 1.8 x 10-5 Calculate [NH3], [NH4+] and [OH-] in a solution originally 0.200 M NH3. 6. If 5.0 moles of O2 and 4.0 moles of NO were entered into an empty 1.0 liter flask, calculate the equilibrium constant if the amount of NO2 found at equilibrium was 1.5 moles. 7. Two moles of NH3 were entered in a 1.0 liter container at 650C. At equilibrium only 71% of the original NH3 was found. Determine the equilibrium constant of the following reaction. 8. If the equilibrium constant for the following reaction is 0.10, determine the final concentration of ICl, if 4.0 moles of I2 and Cl2 were entered initially into an empty 1.0 liter flask. Equilibrium Practice Problems 2. Write the equilibrium expression for each of the following reactions: N2 (g) + 3 H2 (g) 2 NH3 (g) K = [NH3]2 [N2] [H2]3 I2 (s) + Cl2 (g) 2 ICl K = [ICl]2 [Cl2] NO2 (g) NO K = (g) (g) + ½ O2 (g) [NO] [O2]1/2 [NO2] 2. The dissociation of acetic acid, CH3COOH, has an equilibrium constant at 25C of 1.8 x 10-5. The reaction is CH3COOH (aq) CH3COO- (aq) + H+ (aq) If the equilibrium concentration of CH3COOH is 0.46 moles in 0.500 L of water and that of CH3COO- is 8.1 x 10-3 moles in the same 0.500 L, calculate [H+] for the reaction. K = [CH3COO-] [H+] [CH3COOH] Plug in known values and solve. 1.8 x 10-5 = [8.1 x 10-3 moles/0.500 L] [H+] [0.46 moles/0.500 L] [H+] = 1.0 x 10-3 M 9. Indicate the effect of a catalyst, pressure, temperature and concentration on each of the following on: catalyst pressure temperature concentration a. speed presence increase direct relationship direct relationship direct relationship b. posn based on moles based on enthalpy of rxn inverse relationship nothing 10. Given the initial partial pressures of (PPCl5) = 0.0500 atm, (PPCl3) = 0.150 atm, and (PCl2) = 0.250 atm at 250C for the following reaction, what must each equilibrium partial pressure be? PCl5 (g) PCl3 (g) + Cl2 (g) Kp = 2.15 Determine reaction quotient, Q = 0.150 x 0.250 = 0.75 0.0500 Shift to products Make a chart describing relationships in change. PCl5 (g) PCl3 (g) + Cl2 (g) Initial 0.0500 0.150 0.250 Change -x +x +x After 0.0500 -x 0.150 +x 0.250 + x Plug into equilibrium expression and solve for x. 2.15 = (.150 +x)(0.250 + x) so x = 0.0272 (0.0500 –x) Determine Concentrations. PCl5 (g) After 0.0500 -x Substitute 0.0228 PCl3 (g) + 0.150 +x .177 Cl2 (g) 0.250 + x .277 Confirm value of K with these values to check your answer. 11. Calculate Kp for the following equilibrium at 250C if K = 0.042. PCl5 (g) PCl3 (g) + Cl2 (g) K = Kp(RT)- n SO 0.042 = Kp (0.08206x523)-(2-1) Therefore Kp = 1.8 12. The reaction 2 NO (g) N2 (g) + O2 (g) has a value of K= 2400 at 2000 K. If 0.61 g of NO are put in a previously empty 3.00 L vessel, calculate the equilibrium concentrations of NO, N2, and O2. Make a chart describing relationships in change. 2 NO (g) N2 (g) + O2 (g) -3 Initial 6.8 x 10 0 0 Change -2x +x +x -3 After 6.8 x 10 -2x x x Plug into equilibrium expression and solve for x. 2400 = [x] [x] so x = 0.0034 -3 2 [6.8 x 10 -2x] Determine Concentrations. 2 NO (g) After 6.8 x 10-3- 2X Substitute 0 N2 (g) + X 0.0034 O2 (g) X 0.0034 Does this make sense considering the value of K. 7. Using the same reaction as in #6, calculate the equilibrium concentrations of NO, N2, and O2 if the initial concentrations of each species are: [NO] = 0 M, [N2] = 0.850 M, [O2] = 0.560 M. Make a chart describing relationships in 2 NO (g) N2 (g) + Initial 0 0.850 M Change +2x -x After 2x 0.850 M - x Plug into equilibrium expression and solve for 2400 = [0.850 M - x] [0.560 M - x] [2x]2 Determine Concentrations. 2 NO (g) N2 (g) + After 2x 0.850 M - x Substitute 0.014 0.843 change. O2 (g) 0.560 M -x 0.560 M - x x. so x = 0.007 O2 (g) 0.560 M - x 0.553 Does this make sense considering the value of K. 8. The following reaction has an equilibrium constant of 620 at a certain temperature. Calculate the equilibrium concentrations of all species if 4.5 mol of each component were added to a 3.0 L flask. H2 (g) + F2 (g) 2 HF (g) Determine molarity of solutions [4.5 mol / 3.0L ] = 1.5 M of all 3 solutions Make a chart describing relationships in change. H2 (g) + F2 (g) 2 HF (g) Initial 1.5 M 1.5 M 1.5 M Change -x -x +2x After 1.5 –x 1.5 – x 1.5 +2x Plug into equilibrium expression and solve for x. 620 = [1.5 +2x]2 so x = 1.33 [1.5 – x]2 Determine Concentrations. After 1.5 –x 1.5 – x 1.5 +2x Substitute 0.17 M 0.17 M 4.13 M 9. Ammonia undergoes hydrolysis according to the following reaction: NH3 (aq) + H2O NH4+ (aq) + OH- (aq) K = 1.8 x 10-5 + Calculate [NH3], [NH4 ] and [OH ] in a solution originally 0.200 M NH3. NH3 (aq) + H2O NH4+ (aq) + OH- (aq Initial 0.200 M 0 0 Change -x +x +x After 0.2 – x x x Plug into equilibrium expression and solve for x. 1.8 x 10-5 = [x]2 so x = .0018 [0.2 – x] Determine Concentrations. After 0.2 – x x x Substitute 0.2 M .0018 M .0018M 10. The equilibrium constant for the following reaction is 600C is 4.0. Initially, two moles of CO and one mole of H2O were mixed in a 1.0 liter container. Determine the concentration of all species at equilibrium. CO(g) + H2O(g) CO2 (g) + Initial 2.0 M 1.0 M 0 Change -x -x +x After 2.0 – x 1.0 - x x Plug into equilibrium expression and solve for x. 4.0 = [x]2 so x = 0.85 [1.0 – x][2.0 – x] Determine Concentrations. After 2.0 – 0.85 1.0 - 0.85 0.85 Substitute 1.15 0.15 0.85 H2 (g) 0 +x x 0.85 0.85 11. If 5.0 moles of O2 and 4.0 moles of NO were entered into an empty 1.0 liter flask, calculate the equilibrium constant if the amount of NO2 found at equilibrium was 1.5 moles. 2 NO (g) + O2 (g) 2 NO2 (g) Initial 4.0 M 5.0 M 0 Change -2x -x +2x After 4.0 – 2x 5.0 - x 2x Using 1.5 = 2x and therefore x = 0.75 M After 4.0 – 1.5 5.0 – 0.75 1.5 Substitute 2.5 4.25 1.5 K= (1.5)2 = 0.085 (2.5)2(4.25) 12. Two moles of NH3 were entered in a 1.0 liter container at 650C. At equilibrium only 71% of the original NH3 was found. Determine the equilibrium constant of the following reaction. 2 NH3 (g) N2 (g) + 3 H2 (g) Initial 2.0 M 0 0 Change -2x +x +3x After 2.0 – 2x +x +3x Using 2.0 – 2x = 2.0(71%) so x = 0.29 After 2.0 – 2(.29) 0.29 +3(0.29) Substitute 1.42 0.29 0.87 K= (0.29)(0.87)3 = 0.095 (1.42)2 13. If the equilibrium constant for the following reaction is 0.10, determine the final concentration of ICl, if 4.0 moles of I2 and Cl2 were entered initially into an empty 1.0 liter flask. I2 (g) + Cl2 (g) 2 ICl (g) Initial 4.0 M 4.0 M 0 Change -x -x +2x After 4.0 – x 4.0 - x 2x Plug into equilibrium expression and solve for x. 0.10 = [2x]2 so x = 0.546 2 [4.0 – x] Determine Concentrations. After 4.0 – 0.546 4.0 - 0.546 2(0.546) Substitute 3.5 3.5 1.1 14. The reaction of carbon disulfide with chlorine is as follows: CS2 (g) + 3 Cl2 (g) CCl4 (g) + S2Cl2 (g) H = -238 kJ Predict the effect of the following changes to the system on the direction of equilibrium. a. The pressure on the system is doubled by halving the volume. b. CCl4 is removed as it is generated. c. Heat is added to the system. 15. The reaction of nitrogen gas with hydloric acid is as follows: N2 (g) + 6 HCl (g) 2 NH3 (g) + 3 Cl2 (g) H = 461 kJ Predict the effect of the following changes to the system on the direction of equilibrium a. Triple the volume of the system. b. The amount of nitrogen is doubled. c. Heat is added to the system. Sometimes it’s our fault that things are out of balance.